Suppose you are proving that an integer derived in a certain

Suppose you are proving that an integer derived in a certain way is even, and that you are at a point in the proof that you have the integer expressed as 7(4k² + 4k + 1) + 1. How should you rewrite this expression to show that the integer is even?

A) 28k² + 28k + 7 + 1
B) 28k² + 28k + 8
C) 20(k² + k) + 8(k² + k + 1)
D) 2(14k² + 14k + 4)

Solution

option D is correct answer

because given integer is in the form of

7(4k²+4k+1)+1 (where k is some positive integer)

=(28k²+28k+7)+1 (since distributive law)

=28k²+28k+7+1

=28k²+28k+8

by taking 2 common out in the above terms, we get

=2(14k²+14k+4)

=2n ( where n is a positive integer such that n=14k²+14k+4)

we know that every positive even integer is in the form of 2n

so 2n=2(14k²+14k+4) is an even integer

hence (D) is correct